Relationship in between geography of a portable team and also the geography of its profinite conclusion

Suppose $G$ is a portable topological team. We can construct the profinite conclusion of $G$ ; allow is call this $\Gamma$.

My inquiries are:

1) Assuming that we understand absolutely nothing concerning the (initial) geography of $G$ apart from that it is portable, exists anything we can claim connecting the geography of $G$ to the geography of $\Gamma$?

I think the response to this inquiry is "no" given that it appears to me that we generally pertain to $G$ as an abstract team when thinking of creating its profinite conclusion.

2) If not (and also, like I claimed, I think the response to (1) is "no") exists a means to construct a profinite conclusion (or something similar to this) of $G$ that thinks about the geography we currently have?

2019-05-13 04:05:50
Source Share
Answers: 1

In concept, the response to inquiry (1) need to be "no".

When it comes to inquiry (2 ), it appears that a standard building and construction would certainly be to create a "profinite conclusion" of the topological team $G$ by taking the inverted restriction of all ratios $G/N$, where $N$ varies over all limited - index open regular subgroups of $G$. The resulting conclusion $\hat{G}$ will certainly have the building that the canonical function $G\to\hat{G}$ is continual, and also it needs to be in some way global relative to this building.

As an example, allow $\mathbb{Z}_2^\infty$ be the straight amount of definitely several duplicates of $\mathbb{Z}_2$, which is a subspace of the boundless item $\mathbb{Z}_2^\omega$. After that $\mathbb{Z}_2^\infty$ has several limited - index subgroups that are closed (as an example the subgroup of components with an also variety of $1$'s), yet it appears to me that every limited - index open subgroup need to be the junction of an open subgroup of $\mathbb{Z}_2^\omega$ with $\mathbb{Z}_2^\infty$. After that the resulting topological profinite conclusion of $\mathbb{Z}_2^\infty$ should certainly be $\mathbb{Z}_2^\omega$.

2019-05-17 15:20:43